Gerbil Genetics ~ Section 4

strict warning: Non-static method view::load() should not be called statically in /home/egerbilc/public_html/sites/all/modules/views/views.module on line 879.

strict warning: Declaration of views_handler_argument::init() should be compatible with views_handler::init(&$view, $options) in /home/egerbilc/public_html/sites/all/modules/views/handlers/views_handler_argument.inc on line 745.

strict warning: Declaration of views_handler_filter::options_validate() should be compatible with views_handler::options_validate($form, &$form_state) in /home/egerbilc/public_html/sites/all/modules/views/handlers/views_handler_filter.inc on line 589.

strict warning: Declaration of views_handler_filter::options_submit() should be compatible with views_handler::options_submit($form, &$form_state) in /home/egerbilc/public_html/sites/all/modules/views/handlers/views_handler_filter.inc on line 589.

This is the inheritance law which states that the inheritance pattern of one trait will not affect the inheritance pattern of another trait. In Mendel's experiments when he crossed one trait, they always resulted in a 3:1 ratio between the dominant and recessive phenotypes like we saw in our monohybrid crosses, however these ratios altered in his experiments when he mixed two traits together. These types of crosses are known as dihybrid crosses.

Up to now we have dealt with just a singular characteristic, in our case, that of the non-agouti mutation, to help explain the laws of simple inheritance. However we know that although these breeding principles hold true, we also know that there are several other mutations at differing loci that deal with other factors that can effect the overall colour and make things a little complicated, so how would we deal with certain matings that pass on several colour characteristics at different loci, and not just one? Well this is where the Punnett square comes into its own, as this method allows us to manually calculate all the various genetic combination of alleles for any gerbil cross.

In addition to the non-Agouti mutation, there exist several other mutations that are capable of changing the gerbils overall colour. The one we will look at to demonstrate a dihybrid cross will be the pink-eyed dilution mutation. This mutation, like non-Agouti is recessive in nature, and is allocated the symbol 'p'. The mutation when homozygous, will turn a normal agouti's eye colour to red, and dilutes the black pigment in the coat so that it is a rich golden colour, the coat variety being known as the Argente Golden. In a self coat, one with the non-agouti mutation, it will change the coat to a lead grey, and it will dilute the eyes to red, this coat colour variety is known as the Lilac.

If we look at gerbils that will have all the possible combinations of these genes of both the A locus and the P locus, we can see that there are several combinations that make gerbils either Agouti, Argente Golden, Black or Lilac.

Firstly for Golden Agouti there would be...

AAPP AaPP AAPp AaPp

For Blacks there would be...

aaPP aaPp

For the Argente Golden there would be...

AApp Aapp

And finally our Lilac variety would be...

aapp

As you can see, there are four possible genetic combinations of Golden Agouti from a heterozygous 'AaPp' X 'AaPp' mating. When we track these genotypes using the Punnet square we can also work out the probabilities for each genotype of Golden Agouti produced by this cross, and although they will look Golden Agouti, eight out of nine of them will be heterozygous and carry recessive genes, and almost half of them will carry both 'a' and 'p'. The same can be calculated for all the various combinations and is shown in the table on the breeding demonstration.

So firstly let's create our F1 hybrid from homozygous dominant and homozygous recessive gerbils at the A and P locus

Now a dihybrid cross is one where the two individuals that are identically heterozygous at two loci, and we are tracking the two genes from each parent to their offspring. Previously where our gerbil was 'Aa' it could only pass on either 'A' or 'a' to it's offspring. This time the Gerbil that is heterozygous would be 'AaPp', so can pass on 'AP', 'Ap', 'aP' and 'ap'. Using our punnet square we can see how these genotypes interact to produce phenotypes in a 9:3:3:1 ratio, this ratio being typical when two gerbils that are both carrying the same two recessive genes are mated together. Even so, these ratios only represent the chances of each of the offspring being a particular colour, and while over a large number of offspring we could expect one in sixteen being a Lilac coat colour, you cannot guarantee this.

The rules of meiosis as they apply to the dihybrid cross are shown clearly both in Mendel's laws of segregation and laws of independent assortment. For the genes on the separate chromosomes, each allele pair shows independent segregation, and the resulting F1 generation produces heterozygous individuals for the two gene traits, where as the second or F2 generation, which occurs by crossing the members of the F1 generation go on to produce a phenotypical ratio of 9:3:3:1. It is interesting to note that if we look closely at one trait at a time, the two different genes involved are still independently inherited in the usual manner, i.e. a 3:1 ratio.

Mendel concluded from his experiments with dihybrid crosses that different traits are inherited independently of each other. This is the general rule of independent assortment, but today's scientists know that this is only true if the genes involved that effect the phenotype are found on different chromosomes. The exceptions to independent assortment occur when genes appear close to one another on the same chromosome and they are usually inherited as a single unit. This is known as gene linkage and was discovered by the British geneticists Reginald Punnett and William Bateson, shortly after the rediscovery of Mendel's laws.

Now we have looked at some very simple breeding strategies and tracked the genes both on a single locus and at two loci, we are left with crossing a heterozygous individual (F1 hybrid) back to a homozygous individual.

The back cross or test cross were simple methods devised by Mendel to verify the genotype of the F1 hybrid, and can be used for checking the correctness of Mendel's law of segregation (using a monohybrid test cross) and the law of independent assortment (using a dihybrid test cross)

In our monohybrid cross of 'AA' x 'aa' all the F1 generation were 'Aa' (heterozygous Golden Agouti), so let's see what would happen when this F1 is test crossed with the homozygous recessive parent ('aa' Black).

In this cross we are aiming to verify two things. 1) To determine the genotype of the F1 Golden Agouti, and 2) to check the correctness of the law of segregation.

So if we backcross our F1 Golden Agouti with the homozygous recessive Black parent and examine the offspring, we know the recessive Black with 'aa' genotype will produce only one type of gamete, i.e. all with 'a' only, However regarding the F1 Golden Agouti, there could be two possibilities:

A) If the F1 Golden Agouti is homozygous ('AA') it would produce only one type of gametes, i.e, they would all be 'A', and as a result of this, the progeny of the test cross would all be Golden Agouti. 'AA'x'aa' = 'Aa' Golden Agouti

B) If the Golden Agouti is heterozygous with 'Aa' genotype and if Mendel's law of segregation is correct, then the F1 should produce two types of gametes, both 'A' and 'a' in equal proportions. So our test cross progeny should be 50% Golden Agouti and 50% Black or a 1:1 ratio.

As you can see from our breeding demo, the actual test cross agrees with the theoretical expectations and proves that the F1 Golden Agouti is a heterozygous dominant (monohybrid) with 'Aa' genotype and that the alleles segregate during gamete formation.

In our dihybrid cross of Golden Agouti ('AAPP') x Lilac ('aapp') the resulting F1 dihybrid is a double heterozygous Golden Agouti with 'AaPp' genotype. So we test cross this gerbil with the homozygous recessive parent Lilac (aapp). We are aiming to test the correctness of Mendel's law of independent assortment. If this principle holds true, the 'AaPp' dihybrid should produce four types of gametes: 'AP', 'Ap', 'aP' and 'ap' in equal proportions. The recessive Lilac Parent ('aapp') should only produce one type of gametes, that being 'ap'. So it is expected that the maximum possible chance combinations between these gametes should produce four kinds of phenotypes in a 1:1:1:1 ratio.